Question

Express the solution of the given initial value problem in terms of a convolution integral. \(4 y^{\prime \prime}+4 y^{\prime}+17 y=g(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0\)

Short answer

Question: Express the solution of the given linear second-order ordinary differential equation with constant coefficients and initial conditions in terms of a convolution integral. Given: \(4 y^{\prime \prime}+4 y^{\prime}+17 y=g(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0\) Answer: The solution of the initial value problem is: \(y(t) = e^{-t/2}(C_1 \cos(\sqrt{15}t) + C_2 \sin(\sqrt{15}t)) + g(t) * \left(\frac{1}{\sqrt{15}}e^{-t/2}\sin(\sqrt{15}t)\right)\)

Step-by-step solution

Make a note of given information and find the complementary function

We are given the following initial value problem: \(4 y^{\prime \prime}+4 y^{\prime}+17 y=g(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0\) To find the complementary function, we write and solve the homogeneous differential equation: \(4 y^{\prime \prime}+4 y^{\prime}+17 y=0\) The auxiliary equation is: \(m^2+\frac{4}{4}m+\frac{17}{4}=0\) This is a quadratic equation with roots: \(m_{1,2}=\frac{-1\pm\sqrt{-16}}{2}\) Since the roots are complex, we can write the complementary function as: \(v(t) = e^{-t/2} \left(C_1 \cos(\sqrt{15}t) + C_2 \sin(\sqrt{15}t)\right)\)

Using the Laplace Transform to obtain the convolution integral

Now we have to find the convolution integral. The idea is to use the Laplace Transform to find the particular solution of the given differential equation. Let's denote the Laplace Transform as \(\mathcal{L}\) and apply it to the given equation: \(\mathcal{L}\{4 y^{\prime \prime}+4 y^{\prime}+17 y=g(t)\}\) Using linearity properties and differentiation rules of the Laplace Transform, we get: \(4(s^2 Y(s)-sy(0)-y^{\prime}(0))+4(sY(s)-y(0))+17Y(s)=G(s)\) We apply the given initial conditions and simplify the equation: \(4s^2Y(s)+4sY(s)+17Y(s)=G(s)\) Now we solve for \(Y(s)\): \(Y(s)=\frac{G(s)}{4s^2+4s+17}\) We apply the inverse Laplace Transform to obtain the particular solution with the convolution integral: \(y_p(t)=\mathcal{L}^{-1}\{Y(s)\}=g(t) * \mathcal{L}^{-1}\{\frac{1}{4s^2+4s+17}\}\)

Find the inverse Laplace Transform of H(s)

Now we need to find the inverse Laplace Transform of \(H(s) = \frac{1}{4s^2+4s+17}\). To do this, we can use partial fractions decomposition: \(\frac{1}{(4s^2 + 4s + 17)} =\frac{A}{(2s+1)+\sqrt{15}} + \frac{B}{(2s+1)-\sqrt{15}}\) Distributing and solving for A and B gives: \(A = - \frac{1}{2\sqrt{15}}\) and \(B=\frac{1}{2\sqrt{15}}\) Now that we have A and B, we can find the inverse Laplace Transform of H(s): \(h(t) = \mathcal{L}^{-1}\{H(s)\} = \mathcal{L}^{-1}\left\{-\frac{1}{2\sqrt{15}}\frac{1}{(2s+1)+\sqrt{15}} + \frac{1}{2\sqrt{15}}\frac{1}{(2s+1)-\sqrt{15}}\right\} = \frac{1}{\sqrt{15}}\left(e^{-t/2}\sin(\sqrt{15}t)\right)\)

Final solution

Now we have everything we need to express the solution of the given initial value problem in terms of a convolution integral. The general solution will be the sum of the complementary function and the particular solution: \(y(t) = v(t) + y_p(t) = e^{-t/2}(C_1 \cos(\sqrt{15}t) + C_2 \sin(\sqrt{15}t)) + g(t) * \left(\frac{1}{\sqrt{15}}e^{-t/2}\sin(\sqrt{15}t)\right)\) The solution of the initial value problem, in terms of a convolution integral, is: \(y(t) = e^{-t/2}(C_1 \cos(\sqrt{15}t) + C_2 \sin(\sqrt{15}t)) + g(t) * \left(\frac{1}{\sqrt{15}}e^{-t/2}\sin(\sqrt{15}t)\right)\)

Key concepts

Initial Value Problem

In the realm of differential equations, an initial value problem involves solving a function from its derivatives coupled with the provision of initial conditions. Consider a spring bouncing with a force applied at various moments in time. To predict its motion, we need not only the physical law that governs it (the differential equation) but also the starting position and velocity (initial conditions).

In our example, the initial conditions are both zero, meaning the function’s value and its first derivative are known at time zero. These pieces of information heavily steer the course of the solution, much like how a pilot adjusts controls for a smooth flight takeoff. These initial conditions are integral and can vastly alter the results, highlighting their importance in accurately mapping the behavior of the equation over time.

Laplace Transform

Diving into the world of mathematical transformations, the Laplace Transform is akin to a magical tool that simplifies complex functions typically involving time, by swapping them into a more manageable, algebraic form. It’s like translating an intricate novel into a simplified summary without losing the essence of the story.

Applied to our problem involving the function g(t), it transforms the unruly differential equation into an algebraic equation in the domain of 's'. This process streamlines solving the equation as it’s often easier to handle algebraic terms rather than differential ones. Notably, it’s essential to keep in mind that the Laplace Transform is reversible, giving us a power-packed pair that delicately bridges the complex and simple realms.

Complementary Function

Imagine a puzzle - intricate and uniquely shaped. This puzzle is our differential equation. To solve it, we look for pieces that fit: these are complementary functions, solutions to the associated homogeneous differential equation, where the external forces, or in our case g(t), are absent. It's a piece of the entire solution that reflects the system’s natural behavior, undisturbed by external influences, much like the tranquil baseline of a heart monitor before any activity.

In our scenario, to obtain the complementary function, we resolved the homogeneous equation, gracefully detaching all external factors, leaving just the internal dynamics at play. It reveals the inherent resonances of the system embedded in its parameters based on its roots, whether real, repeated, or complex, manifesting in various natural motion rhythms.

Homogeneous Differential Equation

A homogeneous differential equation is one that stands solitary without any external terms—completely self-contained, every term is a derivative of the unknown function we're seeking. It's as if we're probing the soul of the system in its purest form, without outside interactions muddying the waters.

In our exercise, we explored this chaste equation, crafting an auxiliary equation that led us to unearth the complementary function - the pure, unforced response of the system. The roots we found, albeit complex, shed light on the oscillatory nature of the solution, reflecting how, even in isolation, this system has a heartbeat, characterized by a particular rhythm and damping effect, echoing the physics beneath.